# Indefinite integration by parts

I was given an excersice to solve wich asks you to prove:

$$\int_0^1 f(r) r dr = 0$$

knowing that:

$$\int_0^1 f(t) dt = 0$$

After doing integration by parts I ended up with:

$$\int f(r) r dr = (\int f(r)dr)r- \int\int f(r)drdr$$

My question is, how should I proceed in order to evaluate the integral with my integration limits between 0 and 1. I tried with Barrow, but I don't know what to do with the double integral and its limis

• "I was given an exercise to solve... " Where is this exercise from?
– user587192
Dec 11, 2018 at 18:57
• "I don't know what to do with the double integral and its limis" No, you don't have a double integral.
– user587192
Dec 11, 2018 at 18:58
• Suppose $F'(x)=f(x)$. Integration by parts says that $$\int_0^1f(x)x\ dx = F(x)x\mid_{0}^1-\int_0^1F(x)dx.$$
– user587192
Dec 11, 2018 at 19:02

You can't prove it, since it is false. If $$f(x)=\sin(2\pi x)$$, then $$\displaystyle\int_0^1f(x)\,\mathrm dx=0$$, but $$\displaystyle\int_0^1x f(x)\,\mathrm dx=-\frac1{2\pi}\neq0$$
It does not work for a linear function: $$\int_0^1 \left(x-\frac12\right)dx=\left(\frac{x^2}{2}-\frac x2\right)|_0^1=0, \ \ \text{but}\\ \int_0^1 \left(x-\frac12\right)xdx=\left(\frac {x^3}{3}-\frac{x^2}{4}\right)|_0^1=\frac1{12}\ne 0.$$